Question

How many units cells are there in $1.00\text{g}$ cube shaped ideal crystal of $AB$ (MW = 60 u) which has a $NaCl$ type lattice.
(Take, $N_A=6\times {10}^{23}$)

• A. $60.2\times {10}^{23}$
• B. $1.00\times {20}^{22}$
• C. $2.50\times {10}^{21}$
• D. $6.02\times {10}^{24}$

Answer 1

The correct option is C $2.50\times {10}^{21}$

$\text{The molar mass of}AB\text{crystal is}=60\text{g/ mol}$
So, $60\text{g}$ of $AB$ will contain $=6\times {10}^{23}$ atoms
Now $1\text{g}$ of $AB$ will contain
$=\frac{6\times {10}^{23}}{60}=0.1\times {10}^{23}$ atoms
Again, since $AB$ forms a $NaCl$ type lattice, $4$ atoms of the crystal lattice forms $1$ unit cell.
So,
$0.1\times {10}^{23}$ atoms will form
$=\frac{0.1\times {10}^{23}}{4}=0.025\times {10}^{23}\text{unit cells}=2.50\times {10}^{21}\text{unit cells}$
So the unit cells present in a $1.00\text{g}$ cube shaped ideal crystal of $AB$, which has a $NaCl$ type lattice, is $2.50\times {10}^{21}$ unit cells.